51nod 1445 变色DNA（最短路变形）

http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1445

题意：

思路：

挺好的一道题目，如果$colormap[i][j]$为'Y'，那么这条边的代价就是前面Y出现的次数。也就是说前面必须得都破坏了这样才能轮到这条边，这样一来跑一遍最短路即可。

 #include<iostream>
 #include<algorithm>
 #include<cstring>
 #include<cstdio>
 #include<sstream>
 #include<vector>
 #include<stack>
 #include<queue>
 #include<cmath>
 #include<map>
 #include<set>
 using namespace std;
 typedef long long ll;
 const int INF = 0x3f3f3f3f;
 const int maxn=+;
 const int mod=1e9+;

 int n;
 int tot;
 char s[];
 int head[];
 int d[];
 bool done[];

 struct node
 {
     int v,w,next;
 }edge[maxn];

 struct HeapNode
 {
     int d,u;
     HeapNode(){}
     HeapNode(int d, int u):d(d),u(u){}
     bool operator < (const HeapNode& rhs) const
     {
         return d>rhs.d;
     }
 };

 void addEdge(int u, int v, int w)
 {
     edge[tot].v=v;
     edge[tot].w=w;
     edge[tot].next=head[u];
     head[u]=tot++;
 }

 void dijkstra(int s)
 {
     priority_queue<HeapNode> Q;
     for(int i=;i<n;i++)  d[i]=INF;
     d[s]=;
     memset(done,,sizeof(done));
     Q.push(HeapNode(,s));
     while(!Q.empty())
     {
         HeapNode x=Q.top(); Q.pop();
         int u=x.u;
         if(done[u])  continue;
         done[u]=true;
         for(int i=head[u];i!=-;i=edge[i].next)
         {
             int v=edge[i].v;
             if(d[v]>d[u]+edge[i].w)
             {
                 d[v]=d[u]+edge[i].w;
                 Q.push(HeapNode(d[v],v));
             }
         }
     }
 }

 int main()
 {
     //freopen("in.txt","r",stdin);
     int T;
     scanf("%d",&T);
     while(T--)
     {
         tot=;
         memset(head,-,sizeof(head));
         scanf("%d",&n);
         for(int i=;i<n;i++)
         {
             int cnt=;
             scanf("%s",s);
             for(int j=;j<n;j++)
                 if(s[j]=='Y')  {addEdge(i,j,cnt);cnt++;}
         }
         dijkstra();
         if(d[n-]==INF)  puts("-1");
         else printf("%d\n",d[n-]);
     }
     return ;
 }